text:
Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag?
For an unpowered glider, can we show through simple geometry involving force vector diagrams, rather than mathematical formulae or graphs, that the best still-air glide ratio is achieved at the angle-of-attack that gives the maximum ratio of Lift to Drag?

That's a pretty tall order for engineering types lol.

Highly related: https://aviation.stackexchange.com/q/80731/4108

This one is also related, but a bit different-- https://aviation.stackexchange.com/questions/3610/how-are-the-glide-polar-and-l-d-ratio-charts-related?rq=1

Re "Highly related: https://aviation.stackexchange.com/q/80731/4108 -- had not come across that one before, and probably wouldn't have posted this one if were aware of it. Arguably not quite a duplicate because of the specification not to use formulae, but very close; answers to the present question will also fit that one.   I would consider deleting the present question if there weren't already answers to it (other than my own).

We could consider "migrating" all answers to this question to the other one (https://aviation.stackexchange.com/questions/80731/why-is-the-l-d-ratio-numerically-equal-to-the-glide-ratio ) -- if all posters wish to post their answers there and delete them here, I will delete the question.  However since we have an answer here with three upvotes already, that might not be a palatable solution to all...

Let's start with a Lift/Drag/Weight diagram, and draw the velocity vector as well:

Since no thrust is involved, Lift+Weight+Drag must sum to zero. Let's shift the origin of these vectors around a bit:

Now, let's finish the velocity triangle, drawing a (horizontal distance), and b (vertical distance). The ratio between a and b is by definition the glide ratio.

The key idea here is to notice that the triangles L-W-D and V-a-b are similar, that is, they are the same triangle up to a scale.
The similitude goes: W <-> v, D <-> b, L <-> a.
This means that the ratio between a and b is the same as the ratio between L and D. The latter is of course the Lift-To-Drag ratio.
So, in any condition, the glide ratio is the same as the Lift-To-Drag ratio.
It then follows that the best glide ratio is achieved for the best Lift-To-Drag ratio.

Please ignore the unrealistically high angle of attack, I used the first plane image I found on google search.

your diagrams are excellent.  Now, what is moving the glider horizontally? Consider an iceboat, being pushed by the wind from the side (glider sinking), yet moving forward faster than the wind (lift).  Vertical drag (caused by altitude loss) is glider fuel.  V min sink may have a lot to do with air pressure underneath the airfoil assisting lift, but pushing the plane forward more slowly.  100% agree with your rate of descent/ horizontal speed triangle.  This is where Vbg is proven more efficient. (Your glider nose should be below the horizon).

The answer is perfect, exactly the same points that I was making, just nicer diagrams and more concise language.  Obviously there are some unspoken assumptions built in-- for example, still air (as specified in the question).  PS @RobertDiGiovanni I'm sure we can find some examples of an aircraft in an unpowered glide with a nose-up pitch attitude though it may not be the norm for efficient sailplanes.

Not to worry, the lift vectors are correct.

You might consider also posting this answer here https://aviation.stackexchange.com/q/80731/34686 .  You might initially get a downvote or two for posting the same answer twice but in the long run it is a nice concise answer and likely will be appreciated wherever it goes.

First a caveat-- please keep in mind that this answer is focussed on steady-state linear gliding flight.  Throughout this answer, when we talk about what happens when we change the angle-of-attack and L/D ratio, we're assuming that we've waited long enough for the glider to settle back into equilibrium at whatever airspeed is appropriate for the new angle-of-attack and L/D ratio.
All three of the vector diagrams below represent the exact same physical situation-- sustained, steady-state gliding flight along a linear flight path in still air. We've represented the aerodynamic forces in black, and the gravitational force-- the weight vector-- in green. The left-hand and right-hand diagrams are easier to understand than the middle diagram, because the force vectors are arranged into a closed polygon (a triangle in this case), which shows us instantly that the net force is zero. On the right-hand diagram, we've also included the airspeed vector, and its horizontal and vertical components.

Lift is defined as acting at right angles to the relative wind, i.e. the undisturbed apparent airflow created by the aircraft's path of travel through the airmass.  I.e. lift is defined as acting at right angles to the flight path through the airmass.  Drag is defined as acting parallel to the relative wind, i.e. parallel to the aircraft's path of travel through the airmass.
Note that K is the glide angle, and note that on the right-hand diagram, the triangle formed by the force vectors and the triangle formed by the velocity vectors are similar triangles.  This is why the still-air glide ratio -- the ratio of horizontal speed to sink rate in still air -- is identical to the ratio of Lift to Drag.
The still-air best glide ratio -- i.e. the best ratio of horizontal distance travelled to altitude lost in still air -- must coincide with the best ratio of Lift to Drag.
Note that in a steady-state glide, Lift is less than Weight by the same proportion that horizontal speed is less than airspeed.  For many calculations, it is a close enough approximation to assume that Lift and Weight are the same, and that horizontal speed and airspeed are the same.  However, our vector triangle is very sensitive to this distinction -- overlooking it forces our vector triangle to collapse into a straight line, which forces the sink rate to be zero and gives us a glide angle of zero, i.e. a glide ratio of infinity.
Also note that the geometry of the triangle is such that the maximum Lift to Drag ratio also coincides with the minimum Drag value, and the maximum Lift value.
Note that the minimum Drag value does not coincide with the minimum value of the Drag coefficient.  The Drag coefficient is minimized at very low angles-of-attack, i.e. at very high airspeeds.  So we need to be careful about what we mean when we say we want to minimize Drag in order to maximize the glide ratio.  This is only true when we are speaking about Drag as an actual force, not as a coefficient.  The same is true for Lift.  While (for reasonably high L/D ratios) Lift varies much less than Drag as we change the L/D ratio, it is still true that in gliding flight, Lift is slightly higher at the point of max L/D ratio than at any other L/D ratio.  But the Lift coefficient is not maximized at the point of max L/D ratio-- rather, it is maximized near the stall angle-of-attack.  Yet in spite of all this, it is an arithmetic fact that when we maximize the L/D ratio, we also maximize the ratio of Lift coefficient / Drag coefficient -- even though we are not individually maximizing either the Lift coefficient or the Drag coefficient.
Warning: do not mistake the right-hand diagram for an indication that to minimize the sink rate, we simply need to minimize the drag vector.  The force vector triangle and the airspeed vector triangle are geometrically similar, but  they are not identical.  As airspeed changes, the entire airspeed vector triangle scales up or down in size, while the force vector triangle is locked in size by the fact that the Weight vector is constant.  If we miss this relationship, we may tend to assume that the minimum sink rate airspeed must coincide with the airspeed for minimum Drag, which is also the airspeed for the maximum L/D ratio.  This is not the case-- as we'll see below, the airspeed for the minimum sink rate is lower than the airspeed for minimum Drag, which is also the airspeed for the maximum L/D ratio.
Every possible angle-of-attack of the wing is associated with a specific lift coefficient, a specific drag coefficient, and a specific Lift / Drag ratio. If we ignore effects due to changes in Reynolds number associated with changes in airspeed, we can make the approximation that the same angle-of-attack of the wing will always be associated with the same lift coefficient, drag coefficient, and Lift / Drag ratio regardless of whether the aircraft is heavy or light-- a change in weight is expressed as a change in the airspeed associated with any given angle-of-attack, but no change in the lift coefficient, drag coefficient, or Lift / Drag ratio associated with any given angle-of-attack.  If we increase weight, any given L/D ratio-- including the maximum L/D ratio-- will now occur at a higher airspeed.  The increase in airspeed will be proportional to the square root of the increase wing loading.
If the still-air best glide ratio occurs at the same angle-of-attack that gives the best L/D ratio, then it clearly follows that we can minimize the sink rate by increasing the angle-of-attack to slow down slightly, even if this degrades the the L/D ratio and the glide ratio a bit.  In other words, if we fly more slowly down a slightly steeper glide path, we can minimize the sink rate.  Obviously there is a limit to this process-- as we continue to increase the angle-of-attack and degrade the L/D ratio, then at some point well before we reach a full-blown stall, we'll reach a point where the increased steepness of the glide path is too much to be offset by the decreased airspeed, and the sink rate will start to increase again.
In fact, the highest L/D ratio occurs at the angle-of-attack that gives the highest ratio of Cl/Cd, while the minimum sink rate occurs at the angle-of-attack that gives the highest ratio of (Cl^3) / (Cd^2), which is also the highest ratio of ((L^3) / (D^2)) / airspeed.  It's clear that when we want to minimize the sink rate, we gain more by maximizing the lift coefficient than by minimizing the drag coefficient, so we should choose an angle-of-attack that is somewhat higher than the angle-of-attack for the best L/D or Cl/Cd ratio.  This means that the airspeed for the minimum sink rate is always slower than the airspeed for the best glide ratio in still air.
Vector diagrams like the ones included in this answer are surprisingly rare in books about soaring flight.  Materials published by the FAA sometimes attempt to include such a diagram, but they are invariably badly botched, with the length of the various force vectors altered so that they could not possibly form a closed vector triangle, thus eliminating any possibility for the reader to understand the relationship between the force vectors and the glide ratio.  (One might optimistically hope that they are trying to make the point that a glider is not always flying in a straight line with no acceleration?)  (See for example figure 3-1 on page 3-2 of the FAA's "Glider Flying Handbook" (2013) -- here they've even included a Thrust vector!)  Even a theory-rich book like "New Soaring Pilot" by Welch, Welch, and Irving (3rd edtion, 1977) is entirely lacking in any presentation of the Lift-Drag-Weight diagram for steady-state gliding flight.  However, see page 233 of that book for a "polar curve" clearly showing that the max L/D ratio occurs at a higher airspeed than the minimum sink rate.
A couple of additional notes--

Our vector triangles also show that the maximum value of (horizontal speed / sink rate) and the maximum value of (airspeed / sink rate) both occur exactly at the maximum L/D ratio, so we can also find the airspeed for maximum glide angle by looking at the plot of (airspeed / sink rate) versus airspeed and picking the airspeed where (airspeed / sink rate) is maximized or (sink rate / airspeed) is minimized.  Naturally, this will not be the airspeed for minimum sink rate.  The airspeed for minimum sink rate will clearly always be lower than the airspeed for minimum (sink rate / airspeed).

Occasionally one encounters a suggestion that the best L/D speed or the minimum sink rate speed has something to do with the point where Induced Drag is minimized.  The vector diagrams above demonstrate that as long as Weight is fixed, when we maximize L/D, we also minimize Drag.  So at the best glide-ratio airspeed, where the L/D ratio is maximized, Drag is minimized.  Since Induced Drag decreases with airspeed and Parasitic Drag increases with airspeed, it follows that at the point of max L/D -- i.e. at the point of minimum Drag -- half the drag must be Induced Drag and half the drag must be Parasitic Drag.  This is the only point where Drag can be at a minimum.  Clearly, the point of best L/D is cannot be tied to a minimum in Induced Drag or a maximum in Lift / Induced Drag.  The ratio of Lift / Induced Drag is highest at very low angles-of-attack, i.e. at very high airspeeds.

For more on L/D ratios and gliding flight and powered flight, see this related A.S.E. answer, and other linked answers-- 'Gravitational' power vs. engine power

Comments are not for extended discussion; this conversation has been moved to chat.

This is a lot of text for a question asking for "simple geometry"...

@Sanchises -- the question is fully answered by the end of the fifth paragraph (third paragraph under the diagram), the rest addresses other topics of related interest.  Do you think the answer would be improved by simply deleting everything after the end of the fifth paragraph?

@quietflyer I think it would but I thought It'd be rude to suggest :) Especially as you asked the question yourself, you probably know best what you wanted to know, and my preference for short & concise answer is definitely a personal one. (It wasn't my downvote here)

related-- derivation of why min sink occurs at max Cl^3 / Cl^2 -- https://aviation.stackexchange.com/a/82201/34686


text:
Do the soaring and/or ballooning communities in the US recommend 123.5 as the preferred frequency for air-to-air and air-to-ground use?
Does the soaring community (e.g. the Soaring Society of America) or the ballooning community in the US recommend 123.5 as the preferred frequency for air-to-air and air-to-ground use by glider pilots or balloon pilots speaking to other glider or balloon pilots or ground support crew, as opposed to the other "aviation support" frequencies, such as 121.95, 122.775, and 122.85?
Context: In the US, what radio frequency or frequencies should be used by glider pilots and balloon pilots for air-to-air and air-to ground communications? as opposed to the other "aviation support" frequencies?

FAA in AC 90-50, allocates the related frequencies as follows:

121.950: Aviation Instructional and Support
122.750: Aircraft Air-to-Air
122.775: Aviation Instructional and Support
122.850: Special Use and Aviation Support on Non interference Basis
123.025: Helicopter Air-to-Air
123.300: Aviation Support
123.500: Aviation Support

Actual use

Glider schools use 123.3 and 123.5 MHz,

balloons re-use the same frequencies for air-to-air and air-to-ground (chase vehicle) traffic.

Aircraft air-to-air frequency is 122.75 MHz.

Gliders
It is confirmed this is the practice, e.g. for Greater Boston Soaring Club:

The frequencies of 123.3 and 123.5 MHz are assigned to flight schools
and glider operations.  The soaring community uses 123.3 and 123.5 for
glider to glider as well as glider to ground communication.  123.5 is
used when 123.3 is too busy. 123.3 is also used as the "company"
frequency for various FBOs and flight operations.

FAA position for balloons
From FAA Balloon Flying Handbook:

There is confusion among pilots regarding frequencies that may be used
from air-to-ground, balloon-to-chase crew, for instance.
Many balloonists use 123.3 and 123.5 for air-to-ground (pilot-to-chase
crew), as these frequencies are for glider schools and not many
soaring planes are in the air at sunrise.
Since all users of the airwaves must have an ID or call, ground crews
may identify themselves by adding “chase” to the aircraft call sign.
For example, the chase call for “Balloon 3584 Golf” would be “3584
Golf Chase,” or perhaps simply “84 Golf Chase.”
The air-to-air frequency is 122.75. Remember that everyone in the air is using this frequency; transmissions should be kept brief. A
balloon pilot trying to contact a circling airplane would try 122.75
first.

AOPA
For AOPA:

Legally, for air-to-air communications between private, fixed-wing
aircraft, there is just one authorized frequency: 122.75 MHz. For
general aviation helicopters: 123.025 MHz. Gliders and hot air
balloons share 123.3 and 123.5 MHz.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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text:
Is this wording that appears in the legend of the "Chart Supplement" re "airspace" actually used in any entry in the Chart Supplement?
On page 26 of the legend for the February 25 2021 issue of the "Chart Supplement" for the NE region of the US, in reference to airports where Class D airspace converts to surface-level Class G airspace when the tower is closed, it is stated that the "Airspace" entry in the Chart Supplement will typically read

AIRSPACE: CLASS D svc ‘‘times’’ other times CLASS G with CLASS E 700´
(or 1200´) AGL & abv

I've never noticed any entry in the Chart Supplement that refers to the status of the airspace at 700' or 1200' AGL.  For example, on page 30 of the same Chart Supplement, for the "IGOR I SIKORSKY MEML" airport (KBDR) at Bridgeport CT, the "Airspace" entry in the Chart Supplement simply reads

AIRSPACE: CLASS D svc 1130–0300Z‡; other times CLASS G

The VFR sectional chart shows us that the airspace at this airport will become Class E above 700' AGL when the Class D airspace is not in effect, but there is no reference to this in the "airspace" entry in the Chart Supplement.
So is the notation (quoted above) that appears in the Chart Supplement legend ever actually used in any entry in the Chart Supplement?
Does the legend refer to a new change in notation that has simply not been implemented yet?
Or is it a holdover from an older notation format that has since been simplified?

FWIW, searching for "with class E" in the Chart Supplement PDFs finds nothing apart from the three occurrences in the legend. I have no idea if it's a holdover from the past or something new, but a holdover seems more likely.

Link isn't working-- guess FAA doesn't maintain links for past chart supplements-- but you can see the same thing on the current one, downloadable from this page https://www.faa.gov/air_traffic/flight_info/aeronav/Digital_Products/dafd/

Is this wording that appears in the legend of the "Chart Supplement"
re "airspace" actually used in any entry in the Chart Supplement?

No, it isn't.  I would have noticed it by now, have been looking through the Chart Supplements a lot lately.  Have checked multiple airports where this language ought to appear in the AIRSPACE entry in the Chart Supplement.  Examples:
Chart Supplement for Northeast region, effective beginning January 27 2022:
p.32
BRIDGEPORT
IGOR I SIKORSKY MEML (BDR)(KBDR)
AIRSPACE: CLASS D svc 1130–0300Z‡; other times CLASS G.
p.34
DANBURY MUNI (DXR)(KDXR)
AIRSPACE: CLASS D svc 1200–0300Z‡; other times CLASS G.
p.36
GROTON (NEW LONDON)
GROTON–NEW LONDON (GON)(KGON
AIRSPACE: CLASS D svc 1200–0300Z‡; other times CLASS G.
p. 37
HARTFORD–BRAINARD (HFD)(KHFD)
AIRSPACE: CLASS D svc 1100–0500Z‡; other times CLASS G.
Note the lack of any reference to the floor of the overlying Class E airspace that would be in effect when the Class D airspace is closed.
Re the suggestion "Does the legend refer to a new change in notation that has simply not been implemented yet?"  -- that seems extremely unlikely, as this language has been implemented in the Chart Supplement LEGEND, but not in any of the actual AIRSPACE entries in the Chart Supplement text, for nearly a year now since this question was originally posted, and possibly for much longer.
Re the suggestion "Or is it a holdover from an older notation format that has since been simplified?" -- that just doesn't make sense and is undoubtedly not the case.  There's no reason this useful information would be deleted from the AIRSPACE entries in the Chart Supplement-- it doesn't take up that much extra room.
It would make sense that this content might be shortened on the corresponding airspace labels on the actual sectional charts, where there's obviously a high priority on reducing clutter, but I've seen a lot of sectional charts published over quite a few years and am certain that this language (in the longer form) never appeared there.  (No implication is intended that there's any actual requirement that the phrasing of the AIRSPACE descriptions in the Chart Supplement should exactly match the corresponding labels for part-time surface-level controlled airspace on the actual sectional charts.)
So, this language appears to  reflect content that was planned or considered at one time for the Chart Supplement, but never yet actually implemented.  If adopted, this convention would be a useful supplement to the airspace labels on the sectional charts themselves, which say nothing about the overlying airspace.  In some cases it might be easier for a pilot to obtain this information by checking the AIRSPACE descriptions in the Chart Supplement (if they had been revised to the longer form description), than by checking the sectional chart itself for nearby magenta or blue vignetting.  That's what the Chart Supplement is intended to do-- to provide useful supplemental information.
And even if it is felt that this information is already adequately obtainable from the chart itself, at least the Chart Supplement LEGEND ought to match the actual AIRSPACE entries in the Chart Supplement.
Have submitted a suggestion for correction to either the Chart Supplement LEGEND, or the Chart Supplement AIRSPACE entries, via the on-line portal for such things.

Material quoted in answer (immediately after sentence "Chart Supplement for Northeast region, effective beginning January 27 2022:") could be put in standard "quotes" format --

"and am certain that this language (in the longer form) never appeared there --EITHER""

Chart supplements downloadable from this FAA link -- https://www.faa.gov/air_traffic/flight_info/aeronav/Digital_Products/dafd/

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
what is the actual force causes this issue according to IAS that shows normal descend speed?
When we have tail wind during approach , with respect of Aerodynamic forces that effect on airplane , the whole part of fuselage move forward due to tailwind, also we have about 70kts IAS (in general aviation) . In this situation we unable to make normal descend as the same as no wind condition . Now my question is : what is the actual force causes this issue according to IAS that shows normal descend speed?

It's not a force.  It's a velocity that causes the differences.  Assuming you are defining "normal descent" as given glide path relative to the ground, rather than a given descent rate.

If you are just going for a given descent rate and you don't care about the glide path w/ respect to the ground, then the basic premise of the question is flawed; there is no issue at all; the required combinations of angle-of-attack (or pitch attitude) and power setting are identical in the cases with and without the tailwind.  The question could use more clarification as to whether your goal is a given vertical speed, or a given glide path with respect to the ground; and if the former, then the question needs improvement because it is based on a flawed premise.

Related, but probably contains too much detail until the more basic priniciples are thoroughly understood -- https://aviation.stackexchange.com/a/56259/3468

Anyway, will wait till the question is a bit clearer before attempting to convert some of these comments into an actual answer.

Are you asking why you need less power when trying to descend on a given glide slope (with respect to ground) w/ a tailwind, than in no wind, using the same airspeed in both cases?  And in more detail what is the force that causes this difference, or if it's not a force, then what is causing the difference?  Critical to specify what you are holding constant (I'm guessing glide path respect to ground, and airspeed) and what you are allowing to vary (I'm guessing power, and vertical speed, and naturally groundspeed too.)  But it's not clear from the question as it stands at present.

My intention is when I want land on rwy in 10 miles on final , in no wind condition we have special power adjust and pitch attitude to gain about 700 or 800ft per minute to reach 3.00 degrees glide path . but when we have tailwind , what is the actual forces cause more rate of descend and more pitch attitude to gain 3.00 degrees of glide path? with respect of all other parameters have been constant

OK, I understand.  But the question needs to be made more clear, it needs to contain the information embedded in your comment above.  One of my comments was essentially an answer, but I'd rather not post it as such until the question has been made more clear.  But maybe your needs have been satisfied at this point; I see an actual answer has been posted now too.

I got my answer in comments but if you want I say again the question clearly .
What is the actual reason causes less rate of descend per nautical mile when we have tailwind?

The 3 degree glidslope remains fixed relative to the earth, so a higher rate of descent will always be required to stay on the slope if you increase speed.  Conversely,  less rate of descent is needed at slower speeds.
It just makes sense if you think about it, but if you need a practical demonstration try this:  Draw a diagonal sloping line from corner to corner on a piece of paper.  Square it up in front of you, and place your finger at the corner furthest from you.  Now pull the paper left or right, and move your finger toward you to trace the line.  If you move the paper very slowly sideways you can move your finger slowly.  If you move the paper quickly you will need to move your finger quickly.
This is the effect of wind, it changes your groundspeed.  With a tailwind you are moving more quickly over the ground, so you have to come down faster or you will go above glidepath.
Does this help?

So its depend on groundspeed and not a exterior force , when the groundspeed becomes higher, rate of descend must be increases to maintain 3.00 degrees of glide path . Its done . thanks for your explanation


text:
When was the first THS (Trimmable horizontal stabilizer) used and on what aircraft?
When was the first THS (Trimmable horizontal stabilizer) used and on what aircraft? Has any UL or GA trimmable stabilizer?

Do you consider all moving tails as a form of THS?

I mean combination of  trimmable stabilizer to provide stability and elevator used for maneuvering.

If the question is understood to encompass any form of horizontal stabilizer which may be adjusted in flight (with or without a separate movable elevator), then the answer would appear to be Sir George Cayley's 1849 full-sized glider which briefly took flight-- see http://www.flyingmachines.org/cayl.html.  The Wright brothers also used all-moving horizontal surfaces on their early gliders and airplanes, but located in front, as canards.
However, a comment has clarified that the question is meant to only encompass a horizontal stabilizer which may be adjusted for trimming the aircraft, with an attached elevator for pitch control inputs.
According to page 9 of the book "Aces Wild" by Al Blackburn (see this Googlebooks link), the first use of an in-flight-adjustable horizontal stabilizer was in 1929 by Clarence Gilbert Taylor in his B-2 "Chummy", which eventually evolved into the Taylor Cub and then the Piper Cub, all of which used the same stabilizer trim system.  Naturally, all these aircraft featured an elevator attached to the adjustable horizontal stabilizer.
The same source says that many old Waco biplanes used the same feature.  In this photo of a 1936 Waco biplane, the slot associated with the adjustable horizontal stabilizer is clearly visible.

It would appear to be the Boeing 707/720, followed by the DC-8.
Lots of GA airplanes derive pitch trim and static stability from the stab surface, that is operated manually by a screw jack driven by the trim wheel, and only use elevator for maneuvering inputs (having no trim tab, it just trails when hands off). Significant ones that come to mind are the Cessna 180/85 family, the Piper Cub/Super Cub/Pawnee family, and the Mooney M20 family.
Modern airliners with trimmable stabs and hydraulic controls use a particular geometry in the elevator control linkage (or by software if FBW) at the control linkage interface between the horizontal stab and and vertical stab that always keeps the elevator actuator linkage neutral position aligned with the stabilizer chord line, so the surfaces are always hydraulically maintained in stab chord alignment in the absence of a control input.
The result is the trimmable surface becomes the entire stab/elevator unit, not just the stabilizer.  As well the control stick neutral position in the cockpit never changes, which takes getting used to when transitioning from an airplane with a elevator tab for trim, where the elevator stick position changes with trim.

It was first used on Spitfire in October/November 1944 to test out the idea, and was then to be used on the 1943-designed Miles M52. The M52 was going to be a jet-powered supersonic aircraft and would have been the world's first supersonic aircraft. It's not well-known because the design was given to Bell as they started design of the rocket-powered X-1, followed by the British government cancelling the project.

The THS (Trimmable Horizontal Stabiliser) is just the Airbus name for the elevator trim. In Boeing planes it's called Stab trim. All planes have a trim on the elevator: the differences you can find is whether the trim rotates the whole elevator or only a small surface in it

No, a THS means that the entire stabilizer is rotated for pitch trim, not the elevator or a small surface on the elevator.

That is a broader sentence on the trim depending on the aircraft, maybe not directly relevant to the question


text:
How attitude, speed, and altitude vary when increasing pitch?
I have been struggling with the following question for a while, would greatly appreciate some help and explanation why.
Scenario:

I'm cruising in my Cessna- cruise attitude 2300 RPM trimmed for 95.

I pitch up, say 1 inch of back pressure, and hold it.

Initially the plane starts climbing and slows down, but what happens next?
How will the plane look in say 5 minutes with me still holding that 1 inch pitch up, what attitude, speed, and altitude will it have?

Are you talking about one inch of back pressure as measured by the back pressure to get that displacement on the artificial horizon or the back pressure to initially get one inch of backward movement of the yoke?

It may interest you to read the well-known aviation book "Stick and Rudder". Chapter 1 deals with the relationship between attitude, speed, weight, and angle-of-attack.

You will fly more slowly and climb, but the amount of change depends on the static stability of the plane. Without knowing that, I cannot be more precise.

Terry- I think he means just what effect does pitch have on the 4 forces.
To answer your question I think he means just any elevator deflection relative to the trimmed cruise position, just to pull the control column 1 inch back relative to cruise meaning the slightest deflection of the elevator upwards.

My attitude improves with pitch and speed...

An inch is not a measurement of pressure.  Do you mean you move the yoke one inch aft of its trim position under the existing circumstances?  If so please modify question to say so.

PS why not try it and see?

PPS after those 5 minutes pass, are you still holding the same back PRESSURE as you initially applied, as best as you can? Or are you still holding the yoke exactly 1 inch aft of its original trim POSITION? If the latter, how do you know? Did you mark the position with a grease pencil or something, on the stem that protrudes from the instrument panel? Did you wrap (and tape) a paper around the stem so that you could make pencil marks? Did you tighten a sheet-metal-clamp vice grips (w/ padded jaws of course) on the stem at trim to show the place where it protruded from the panel?

I've done all those things and they are good ways to explore the effects of change in the POSITION of the control yoke. without regard to PRESSURE.  Also, you can simply tighten your fingers tightly around the stem where it protrudes from the panel to mark the spot, and try not to let them slip.

Or did you mean, you continually apply the same force or PRESSURE (as best as you can) that INITIALLY moved the yoke one inch aft of its trimmed position, without any regard to its CURRENT position? That would be a whole different scenario. So, the question needs clarification before it can be answered The question could lead to a rather interesting set of answers but it needs to be cleaned up a little first so that it is more clear what the actual question is

The point being, as airspeed changes, so does the trimmed position of the control yoke, and so does the force or pressure needed to hold the yoke a given distance away from its current-- or original-- trimmed position. So your question is juggling a lot of variables, without making it clear which are the ones that you intend to hold constant.

Well, you knew to be specific about the plane, the RPM, the airspeed, and the attitude.
Pulling the yoke back without changing anything else will create excess lift, and a climb begins.  This is the only point in the scenario you have excess lift because:  increasing AOA increases drag.
The plane then slows down for 2 reasons.  First a higher AOA means more drag.  Secondly, as the plane pitches up, the gravity vector increasingly contributes to the drag vector.  (A descending aircraft, nose "down", allows gravity to contribute to thrust).
Specifically for a 172, there are 2 outcomes.  In this case a steady climb is likely at a lower airspeed.  If the yoke is pulled harder, the plane continues to pitch up and stall (with more power and/or airspeed it may loop, but I would not bet on it at 90 knots, 75% power from level flight in a 172).
But 5 minutes later?  As you climb, the engine produces less and less thrust.  Assuming static stability (a hallmark of a properly CG balanced 172, checked pre-flight), the rate of climb will decrease until the plane has insufficient thrust to continue climbing.  It will then fly level at its trimmed airspeed.  (If you have sufficient oxygen and no head wind, your ground speed will be higher).
But this provides an insight into what it is doing when it climbs in the first place.  Since the wing is around 4x more efficient than the prop in producing lift, pitching up and "using the prop to climb" does not work very well.  It turns out the mechanism of static stability works as well in a climb as anywhere else.  As the plane pitches up and slows, enough lift is lost to cause the plane to sink (even though the nose is skyward).  Sinking pitches the nose down, which increases speed, which increases lift.  this works at any power setting, from gliding to full power..
We can see, "5 minutes" later, the plane no longer can climb, but still will be staticly stable at its trim speed so... cut the power, and enjoy the glide home at that airspeed (if you can make it), or, at least remember CARB HEAT ON.
And also, consult the POH for that flight envelope chart, and remember, excessive pulling the yoke can exceed G limits as well.

Thanks guys.
So what you are saying is, from our cruise at 95 IAS say 2000 altitude we will enter a climb, our speed will decrease, and we will slowly climb until say at 10,000 feet finally power required curve meets the power available curve, our plane stops climbing (at say the new speed of 80/70),
And speed will increase back to the cruise 95?

Sorry Robert,  not sure Im following you. I understand it will climb and when it reaches power required=power available the climb will stop. But what would the new  cruise IAS be? You are saying it will cruise at its Trimmed airspeed. Wasn’t that trimmed airspeed 95?

No, because you changed trim by pulling the yoke back.

Ohh, I think Im starting to get it. So our new cruise altitude will be higher, speed will be lower: Thank you.

@leha007 well, you're getting comments anyways.  Trim speed will not change, rate of climb does. If you go to max altitude, then let go of the yoke, your trim speed returns to 95, and your plane will return to its original altitude, because it has insufficient thrust to do 95 indicated at that altitude.

I would guess that it would take much more than 5 minutes for the plane to stop climbing under the circumstances described in the question, assuming that we are starting the experiment not too high above sea level.

The general answer goes something like this:
Level flight, at it's most basic involves a balance of 3 factors:

How fast are you going, relative to the wind?
What is the angle of attack (AOA) of your wings, relative to the wind?
How dense is the air that you are flying through, based on altitude, temperature, humidity, etc.?

Speed and AOA are variables that you indirectly control, via throttle (by increasing thrust) and yoke (by using elevator to alter pitch) respectively. The thing is, changing one variable (speed, AOA, or density) requires that one or both of the others change to maintain equilibrium. Want to stay in level cruise at a given density altitude after lowering throttle? AOA must increase. If the AOA stays the same, the aircraft will begin to sink because the current density cannot provide enough lift at that combination of AOA and speed.
That explains some of the "Why". Now let's consider your scenario. You have taken throttle out of the equation. Thrust is fixed and cannot be changed. This means that for any change to AOA, density must provide the equilibrium. Unless you are already at your stall speed, a slight increase in AOA without lowering throttle will begin a climb at slower forward speed because this new combination of AOA and speed are producing excess lift at the current air density. You have traded forward speed for extra lifting force. Because the throttle and new pitch are now fixed, the airplane will simply climb at it's new speed and AOA until altitude and temperature lower the outside air density to the new equilibrium point. At this point, the aircraft will once again be configured for level flight at the new density altitude. Also, because the density will change gradually, so too will the rate of climb decrease gradually. It will probably take a good while to fully level out.
ADDENDUM:
The "lift" topic always sparks arguments, because technically, lift can be generated in any direction, depending on the definition of lift you are considering. An aircraft moving through the air on its side, may still create some small amount of force laterally, which opposes gravity, even though the wings may be producing no force at all. A helicopter in hover is utilizing all of its accelerated airflow for to counteract gravity, but a change in rotor disc tilt divides that total force between lift (opposing gravity) and thrust (opposing drag). An aerobatic airplane hanging on its prop is utilizing accelerated airflow in much the same way. An aircraft flying upside down is creating lift that is simultaneously against the pull of gravity AND toward the bottom of the aircraft.
Additionally, Bernoulli's principles work equally well in any orientation. An asymmetrically cambered airfoil mounted vertically will still create an area of lower pressure on one side than on the other, and is still capable of altering angle of attack to adjust deflection. Is this lift? The simple physics definition would say "No" because it is not opposing gravity. An aerodynamic analysis of the airfoil would say "Yes", the airfoil is creating lift according to its orientation, which in this case, affects only yaw.
It would be nice if the entire world could agree that Lift is the sum of all forces that act in opposition to gravity, and Thrust is the sum of all forces that counteract drag - regardless of whether these forces are provided by airfoil or prop (which is still an array of airfoils!). Unfortunately, for now, clarification is often required when discussing this problem in polite company.
In closing, I would like you to consider this: in your scenario, the reason that your airplane slows down is because, much like the helicopter, your prop disc is being tilted, dividing its power between lifting and thrusting. The drag on the airplane remains, so the added burden of lift is what leeches away the speed. The wings, on the other hand, no longer bear the full burden of opposing gravity, so the lift that they are producing DOES become less, but only because they are now sharing that load with the prop.

Thanks guys.
So what you are saying is, from our cruise at 95 IAS say 2000 altitude we will enter a climb, our speed will decrease, and we will slowly climb until say at 10,000 feet finally power required curve meets the power available curve, our plane stops climbing (at say the new speed of 80/70),
And speed will increase back to the cruise 95?

Yes, you will climb. Yes, your speed will decrease. 10,000 feet is speculative - it depends on factors that affect the air density and how much trim you have already applied. I doubt that you will climb 8,000 feet from one inch of aft yoke (especially before running out of gas). Yes, when the curves are in equilibrium you will once agin be in level flight.

No, you will not increase speed once you get back to level flight. This will be the new density altitude that supports level flight for this new AOA/Speed combination - the air is thinner, so less thrust, so less excess power to make you climb, right up until you are level and can maintain the new, slower speed (80 or 70 kts, whatever it is).

Got it! Thank you.

And another one to see if I understand, say from cruise i apply back pressure, but this time don’t lock control column and release it. Will the plane climb up, then down then up then down in a series of decreasing oscillations until it is back at the same altitude at 95?

@Aaron Holmes thanks, your explanation re-inforces a concept that is in the process of being improved on this site:  lift is greater than weight in a climb.  A great many people here thought it was less.  At any power/trim setting, a staticly stable plane will pitch up (if too fast) or pitch down (if too slow). The throttle determines angle to the horizon and vertical lift determines rise or sink.  As a 172 pitches up, the vertical thrust component is insufficient to replace the vertical lift (1:4 ratio), and the plane climb angle equilibrates based on thrust.

@RobertDiGiovanni -- Re " lift is greater than weight in a climb. "-- nope. Not in a stabilized steady-state climb. In such a case, lift is less than weight.  See for example https://aviation.stackexchange.com/a/56476/34686.

@quiet flyer: See the top rated answer on the link you posted. In this instance, we are defining lift in the way that a physicist would, as "the sum total of all forces acting against gravity", instead of "the sum total of all forces acting perpendicularly toward the top of the aircraft".

This one will die hard, but it must. First vertical lift is cosine lift, there for the lift must be greater than weight if the plane is pitched up or down.
Remember, using the wing is always more efficient than thrust or drag! Second, any vertical thrust contribution is only 1/4 (for a 172) the vertical lift lost from pitch.  In a "stabilized steady state" all upwards forces = weight, essentially rendering the aircraft "weightless" and thrust acts against drag.  So happy with Aaron, because Rho got in there too.  Rho is also part of the Lift equation, as is increased AOA.

@Robert, how did you arrive at your 1/4 figure? If you are talking about "vertical thrust contribution", then it is not directly comparable to lift and highly depends on pitch. Say, we were level (pitch = 0), 0% thrust component, 100% lift. Pitch up 5°: now the lift loss $1-\cos(5°) \approx 0.4$%, thrust gain $\sin(5°) \approx 8.7$%. If we account for L/D ≈ 10, then there is still 0.87% gain in absolute numbers vs 0.4% loss.

@Zeus at 95 knots 1/4 thrust to lift ratio would actually be a bit generous, and (for the Clarke Y) L/D is much greater than 10 (at optimal AOA).  But I very much like your effort to plug in numbers rather than to argue passionately.  I work backwards from reality, both Vy and Vbg are faster than V min sink because...?  Vy performance is much better than Vx because...?  In the end "the vertical thrust component" is moot for best performance (that you need at maximum altitude).  Thrust only opposes drag (in the direction the plane flies).  The wing does the lifting.

So, the best way to climb is to set AOA at optimal and speed up slightly (Vy).  Increasing altitude brought this home so well, the higher you go, the less you can pitch up (because you can no longer go faster).  The thrust vs airspeed curve for the 172s fixed pitch prop makes it a bit more confusing (and the much slower Vx even possible).

@Robert, it's fairly obvious that wing is the most efficient aerodynamic device. But still, I can't see how you tie it to the prop efficiency. The literal interpretation of your "any vertical thrust contribution is only 1/4 the vertical lift lost from pitch" is the calculation I've done, only with a different result. C172's best L/D is known to be just under 11, so at 95 kt it's probably around 8, and the lower it is, the greater the thrust must be, as well as its vertical contribution.

Well, happy to continue working on it.  A 2400 lb Cessna flies fine with 600 lbs of thrust, but cannot climb on the prop (it falls short 1800 lbs!).  The best way is to use the wing.  In this case, increasing AOA at the same thrust setting causes a climb.  An airspeed reduction from 95 to 70-80 (see drag curve) gives the plane (less drag aerodynamicly and more thrust) the ability to climb, and it does until Rho factor no longer allows it.  Now, if you were on the other end of the power curve, like Langewische, pulling the elevator would cause a higher drag "mushy glide".

But the "vertical lift component" may exist for an F-16, but not for a 172.  Thrust must be used to move the plane, letting the more efficient wing lift it.

As the airspeed decays and then stabilizes at a lower value, lift will end up less in the climb than it was in horizontal flight.  See https://aviation.stackexchange.com/q/40921/34686

Re " in your scenario, the reason that your airplane slows down is because, much like the helicopter, your prop disc is being tilted, dividing its power between lifting and thrusting." -- that's definitely not the primary reason that the airplane slows down when the yoke is moved aft to increase the wing angle-of-attack.  Higher angle-of-attack equals higher drag coefficient from wing which causes the airspeed to decrease.

Part one: generally speaking--

I pitch up, say 1 inch of back pressure, and hold it.

I'm going to assume that you mean that you move the control yoke aft one inch of its original trim position and keep it there, regardless of how the force needed to keep it there changes during any subsequent airspeed changes.  (If you don't re-trim, you'll need to maintain some amount of back pressure on the yoke, but "inches" aren't really a unit of "pressure", and the exact force needed to hold the new control yoke position will change somewhat as the airspeed changes.)

Initially the plane starts climbing and slows down...

Correct

...but what happens next?
How will the plane look in say 5 minutes with me still holding that 1
inch pitch up, what attitude, speed, and altitude will it have?

Well, it depends on whether you were initially flying on the "front side" (fast side) or "back side" (slow side) of the drag curve.
We'll assume for simplicity that thrust remains constant throughout, though that's not exactly true.
If you were flying on the "front side" (fast side) of the drag curve, then slowing down reduces drag, and the airplane will end up in a climb.
If you were flying on the "back side" (slow side) of the drag curve, then slowing down increases drag, and the airplane will end up in a descent.  Be careful-- you are in the "slow flight" part of the flight envelope, and perhaps may be close to the stall speed.
In either case the airspeed will be reduced.
In the first case (the "front side" case), the pitch attitude will certainly end up more nose-high than before you moved the control yoke.  In the latter case (the "back side" case), that may or may not be true.1,2
In either case the angle-of-attack will have increased.  Think of the control stick or yoke as being most fundamentally an angle-of-attack control, not a pitch attitude control.  Pitch attitude is simply the result of angle-of-attack plus climb angle (or minus descent angle.)3
Part two: since you specified that you were initially flying at 95 mph or knots--
You were initially well on the "front side" (fast side) of the drag curve, so you will end up in a climb when the airspeed stabilizes at the new, decreased value.  The pitch attitude will be shifted in the nose-up direction.  (One inch of change of control yoke position isn't enough to "pull through" to the "back side" of the drag curve.)  As for specific numerical values-- why not try it yourself and report back?  But it's safe to say that your airspeed will still be well above Vy, not to mention Vx.
Footnotes:

Note therefore that the title of the question is somewhat misleading-- in the long run, you are not necessarily "increasing pitch" (attitude) with your control input.

There is a third case where the airplane starts on the "front side" of the drag curve and ends up on the "back side".  In this case the flight path could end up being exactly horizontal after the aircraft settles into stabilized flight at the new angle-of-attack and airspeed.  In such a case, the pitch attitude would have shifted in the nose-up direction.  And there is yet another case, starting at a slightly lower initial airspeed, where the aircraft ends up in a descent, but at the same pitch attitude as it started with.

To be complete, we should add "minus wing incidence" to this sentence-- if we assume for simplicity that wing incidence is zero, then the sentence is exactly true.

PS if your meaning was that you apply whatever aft force is needed to keep the yoke one inch aft of its current instantaneous trim position, the answer would remain essentially unchanged, especially "Part two". Because you aren't going to be causing a big enough airspeed change to see much change in yoke position as the airspeed decays a little, while you hold a constant x ounces or pounds or kilos or Newtons of aft force on the yoke. But the yoke will end up a little further aft of the position that you initially pulled it to, if you continue to hold a constant force as the aspd decays.


text:
accelerated turns in helicopter
In flight Manual of Russian Mi-17 helicopter, it is give that ”max bank permitted in accelerated turns is 45 deg up to 1km altitude”. At the same time it also mention max bank permitted is 30 deg up to 1km altitude.
What do we mean by accelerated turns and why difference between max bank permitted values ?

Can you provide a link to the document that you're quoting from?

Sorry Ralph,  I don’t have soft copy with me.
Thanks for time n effort.

What do we mean by accelerated turn

If "we" is intended to refer to the aviation community as a whole, rather than to the authors of the Mi-17 flight manual--
The term "accelerated turn" has no generally recognized meaning in aviation, and doesn't really make sense.  All turns are a form of acceleration, namely, centripetal acceleration.  Meaning that the net force acting on the vehicle cannot be zero in turning flight.1
My interpretation of the content at the top of page 2-5 of the Mil-17 flight manual is that it is trying to convey that there is a certain speed range (120-250 kph) where, if at low altitude and not above normal operating weight, the helicopter may be banked more steeply than otherwise be permissible.  The choice to apply the specific label "accelerated turn" to this particular situation may be somewhat arbitrary, and may not reflect a standard terminology in the rotorcraft context.2
Footnotes:

In relation to winged aircraft (rotorcraft dynamics may be different), one could concoct a definition of "accelerated turn" that meant that the wing was producing more lift than the aircraft's weight, so that the G-loading was greater than one.  If the aircraft was then brought to the stall angle-of-attack while holding the G-loading constant, a stall entered from this condition would produce an "accelerated stall", where the stall speed was greater than the aircraft's normal straight-and-level stall speed.  Conversely, if the wing was producing less lift than the aircraft weight, so that the G-loading was less than one, a stall entered from that condition while holding the G-loading constant would produce a stall at an airspeed lower than the normal 1-G stall speed.  It is possible to turn with a G-loading of less than one, but (at least in the context of coordinated flight rather than "knife-edge" maneuvers where the airflow is striking the side of the fuselage and creating a lifting force) the flight path will always be curving earthward due to insufficient lift.  And therefore such maneuvers may only be sustained for short time scales.  (The "top" part of a steep-banked, exaggerated "wingover" is an example of such a case-- the G-load may drop well below 1 G, and the airspeed may drop below the normal 1-G stall speed, but neither condition can persist for long.)  But even in a turn that was "non-accelerated" according to this made-up definition, the net force on the aircraft would be not be zero and the aircraft would be in fact undergoing a centripetal acceleration-- as well as, in most cases, an acceleration component along the instantaneous direction of the flight path.  So let's throw this made-up definition into the garbage bin.

Thanks to this related answer for bringing this page in the flight manual to our attention.

@RTO-- good point; my footnote was addressing the airplane situation.  I don't have much knowledge about helicopters.  Will edit to clarify.

Thanks quite flyer

I would interpret an accelerated turn in contrast to a steady turn.
A steady turn is one that can be conducted without losing energy -- speed and altitude.  It is fundamentally a power limit.
Turns can also be limited by stall or structural limitations.
If (at a certain flight condition) you turn at greater than the max steady turn rate (but still within stall and structural limits), you will have to bleed off speed and/or altitude to continue the turn.
Of course, the other way to have an accelerated turn (i.e. a turn that is not steady) is to turn at a rate that is less than the steady turn limit (while at full power) -- in that situation, the aircraft would be able to climb or increase speed during a turn.  I don't think this is what they're talking about here.

In other answers on ASE, I'm pretty sure I have seen the word "stationary turn" used to describe what you are calling a "steady turn".  So it seems you are suggesting that what might logically be called a "non-stationary turn", is what the manual is calling a "stationary turn".  But-- in relation to the loads that the helicopter structure can tolerate-- why would it really matter whether the aircraft is losing energy or not?

I'm largely using fixed-wing language when I say steady turn and then extend to accelerated turn.  The structural limit does not care about whether the aircraft is losing energy.  I read all this to mean that the energy maintaining turn limit is 30 degrees, but if you're willing to lose energy, you can structurally go up to 45 deg.

Re your last sentence above-- if that's what the manual intends to convey, I'd say that it's poorly written--

The Mi-17 flight manual shows those limitations on pages 2-4 and 2-5, but under different circumstances.
The normal bank angle limitations have weight and altitude limitations.
The statement of allowed bank angle in accelerated turns states turns up to 45 degrees are allowed in accelerated turns at up to 1000 meters altitude,  airspeeds between 120 and 250 km/h and "normal or subnormal gross weight".
Essentially, the flight manual is telling you the limits during normal flight ops, but also what the limits are if you need to engage in extreme maneuvers, like getting out of the way of a missile about to come up your rear, for example.


text:
Could you use the flaps to recover from a flat spin
If the plane is in a flat spin, what would happen if you put the flaps down to cause the backward turning wing to be pushed down, and the forward turning wing up, to tip it out of the spin.

No.
The chordwise airspeed at the flaps (mounted about twice as close to the CG than the ailerons, rudder, etc.) would be about half the airspeed felt by those already mushy-nearly-ineffective primary control surfaces.  So the roll moment imparted by the flaps would be so slight that it might take hundreds of rotations, and tens of thousands of feet of falling, before significant bank developed.  And then the primary controls are still as mushy as before, so you're no closer to recovering.  See Is it possible to recover from a flat spin?

At 2 rotations per second, the flaps are moving at 20 mph if they are mounted 2.3 feet from the center of rotation. I'm pretty confident most airplanes don't have flaps mounted that far inboard. I also wonder if the extra drag from the flaps could help dampen the yaw rate? I'm not sure but it sounds plausible.

Thanks for doing the math, Chris!  That showed me how to streamline the logic.  As far as yaw drag goes, flaps are small and have little leverage compared to the aft fuselage plus fin.  It would slow yaw rate (so, hey, might as well deploy them) but not by much more.

Thanks! I suspected it wouldn't have much effect, with the flaps so close to the fuselage. I heard that they turn the rudder hard the other way, which, I assume, increases the sideways drag of the tail to flip the nose out of the spin. I have used flat spin to drop a piece of tissue paper, like toilet paper straight down onto a spot of water on the floor. I usually goes straight down. I would hate to be in a plane doing that. The gyroscopic effect would be your worst enemy.

@Camille Goudeseune Is my assumption about how the rudder flips the nose out correct? And, the other method, pulling back on the yoke or stick, which would push the tail down when facing to the back, and upward when facing forward: I assume, if the plane as a whole acts like a spinning top, would get pushed up on the side facing the oncoming air and down on the side facing back. That, with the gyroscopic effect would tip the plane sideways (to the right out of a LH spin) is that correct?

@PeterR.McMahon those two or three new questions would be better asked as fresh questions on this Q&A site, than as conversation.  They would be made clearer by using the terms yaw, pitch, and roll.  And (as often happens to me) while still typing in that question, stackexchange might offer you a suitable answer :-)


text:
What would happen if a plane lost half of its weight instantly?
During the Second World War the RAF had a 10,000 kg bomb, the "Grand Slam", which was dropped from a Lancaster bomber.
I assume that when this was dropped the plane would lurch upwards.
Is there any way of knowing just how big and violent a movement this would have been?
According to Wikipedia, a 'loaded Lancaster' weighed 25,000 kg. I'm not sure if this includes a standard bomb load (approx 6,350 kg) or not, but either way it seems that when the "Grand Slam" was dropped the aircraft would, in an instant, reduce to about half or two-thirds of its previous weight.
Is there any way to estimate how big an upward jump this would cause?

Bombers open spoilers as they drop bombs. It's done exactly to prevent that upward jump, mostly for precision so the next bomb can be dropped without need to re-align the plane. I don't have a link now, but there was a cool video with visible row of small spoilers on the wing that are opening one by one, as the plane loses weight in carpet bombing.

If not balanced properly, you could wind up with something a lot like this (cargo broke loose and slid to the back): https://www.youtube.com/watch?v=4Z7GkV5DqOI

The center of gravity changes too. So there is a need to re-trim the airplane.

@Agent_L -- re your comment-- I have never heard of anything like that.  Can you provide a link?

@quietflyer Unfortunately, I can't.

In steady flight, the lift $L$ generated by the wings equals the loaded weight so that there is no vertical acceleration.  Thus, the lift just before the bomb is dropped is
$$
L = m_1g =  245\,250 \;\textrm{N}
$$
where  $m_1=25\,000$ kg is the loaded mass and $g=9.81$ m/s$^2$ is the acceleration due to gravity.
In the time interval just after the $10\,000$ kg bomb is dropped, the lift remains the same until the lift is reduced to again balance the new mass of $m_2=15\,000$ kg. During that transient period, the Lancaster will accelerate upwards at
$$
a = \frac{L-m_2g}{m_2} = \left(\frac{m_1-m_2}{m_2}\right)g
$$
which works out as an upward acceleration of two-thirds of a "gee":
$$
\frac{a}{g} = \frac{m_1-m_2}{m_2} = \frac{2}{3}
$$
As @sanchises pointed out, the steady increase in the upward velocity results in a decrease in the angle of attack, so that the aircraft will again settle in a steady climb.

There is no upward jump! The discontinuities are with the vertical forces and the vertical acceleration. The vertical speed and the altitude remain continuous when the drop occurs.

Yes, @bogl.  There is no sudden "jump", but a smooth upward motion.

@bogl There is no 'jump' but there is definitely a 'jolt' - or more formally, an extremely high jerk for a very short time as the vertical acceleration goes from zero to $\frac{2}{3}g$.

@sanchises Completely agree. That should be pretty uncomfortable.

Either way, welcome to the community!

I think the decrease in angle-of-attack must be viewed as very temporary-- see my answer

The A7E Corsair had an empty weight of around 19,000 pounds. Its fuel load was 10,500 pounds, and with six armed 2,000 pound MK84 general purpose bombs hung under the wing it had a total weight of around 42,000 pounds. This was maximum takeoff weight, and coming off the cat shot the aircraft response felt lazy and sloppy as you did your clearing turn.
By the time we got to the target, which was a rock in the Aegean called Avgo-Nisi, we were at around 7,000 pounds fuel remaining. That put the aircraft at around 26,000 pounds and a 12,000 pound bomb load still under the wings. We cleared the target with a low pass before climbing up to the pattern altitude burning probably a 1,000 pounds. It was a popular fishing spot and so we wanted to make sure the area was safe before our delivery.
We were in combat formation when we rolled in to our 45 degree dive, arming the pylons. I was on the wing of my lead in a loose formation around 400 to 800 feet away. The formation was a tactical position, and if either one of us got hit during combat operations the other was far enough away to survive. A 45 degree dive looks very steep from the cockpit and we were fast at the release point. Minimum altitude for this kind of run was 2,500 feet AGL to keep you out of the blast radius. We were at 1 G when we released 24,000 pounds of bombs on the rock, 12,000 pounds apiece.
I watched the bombs release from the lead's aircraft as I saw mine drop away in the mirrors. I just shed around half of my total weight. I immediately felt the aircraft "jump up." This is the way I  characterize it, and someone else might call it a jerky motion. It was definitely stronger than the nose tuck you experience in the A7E breaking the sound barrier. The jump wasn't disconcerting, but definitely noticeable. In fact, it was a welcome change to what felt like a lumbering, gas eating configuration.
We pulled out of the dive hard and had our noses high in a turn when I felt the blast wave go through my internal organs. It was a weird feeling, like someone rearranging the position of my stomach. I must have been at 4,000 to 6,000 feet above the detonations and could hear the explosions over all the noise in the cockpit. It was massive.

I like this answer even though some readers may consider it to be too much information. - It is a very clear demonstration that yes, an aircraft does indeed bounce upward upon shedding a large fraction of its weight, but that is far from the most exciting thing that happens on a typical day when using an aircraft in this manner.

I appreciate your comment. Thank you. Using day-to-day experiences of carrier aviation brings home the nature of what we do. I learned best from stories told in briefs, at safety standowns, during NATOPS, or even in the dirty shirt mess. Also, I think there is a common misconception among some, that the bombers today carry less payload than those of old. Lastly, I think valuable information can also be engaging as well, but that does take space. My intention certainly isn't to write for everyone. I, personally, like scanning an answer and if so inclined will read it. Again, thank you.

@Aaron, great to hear from someone who did it and lived to tell the tale.  Thanks.

While 12.000 lb is a pretty sizeable bomb load, it's less than the 10.000 kg bomb load in the question - The Mk84 is 925 kg, so x6 that's merely 5550 kg. Then again, the A7 was carrier-based. A B1-B will carry 24, so a bit over 22 tons. Relatively, the Lancaster dropped 40% of its take-off weight, versus 30% for this A7, so in that sense there's been some regression.

(A new question was recently asked, then closed as a duplicate, linked to this older question.)
Answer: You would have a pitch "phugoid"
Until airspeed decreases, lift is greater than weight
Flight path curves (accelerates) skyward
There is a very temporary reduction in angle-of-attack, until the aircraft's pitch stability dynamics overcome pitch rotational inertia and pitch the nose up into line with the new direction of the flight path, restoring the original angle-of-attack.  This effect may be negligible.
The changing direction of the flight path causes the weight vector to have an aftward component relative to the aircraft's flight path.
Thrust, drag, and the fore-and-aft component of weight are no longer in balance, and airspeed starts to decrease.
Eventually the aircraft will come back into equilibrium in a stabilized climb at a constant airspeed, lower than the original airspeed.  But we might see the aircraft's pitch attitude "overshoot" and "undershoot" the final steady-state pitch attitude several times, as the flight varies between steeper and shallower climb angles, before this happens.  Likewise we might see the aircraft's airspeed "undershoot" and "overshoot" the final steady-state airspeed several times before settling into equilibrium.  Basically what is going on here is that a change in the climb angle drives a change in airspeed which drives a change in the magnitude of the lift vector which drives a change in the climb angle, and round and round it goes till everything settles into balance.
All this is assuming the pilot is making no corrections-- such as to hold a constant pitch attitude-- and the aircraft's pitch stability dynamics are acting to try to hold some given trimmed angle-of-attack, and the aircraft's dynamic pitch stability characteristics are such that a pitch "phugoid" tends to eventually damp out.

Another way to look at this is what would happen if IAS increased 100 mph, which could happen entering a very strong microburst.  This is where building models and gaining experience with tail design and CG/trim condition can provide some key insights.
It is very desirable NOT to have "phugoid" behavior when lift/weight vector comes out of balance. There is a relationship in wing and tail design that is very fundamentally important as the horizontal stabilizer has (with the Lancaster) a long torque arm and end plates, giving it directional stability.  We also must remember pitch change will cause wing center of lift to move forward as well.  Trim condition/CG placement is also very important.
So what will happen?  With the lift/weight vector suddenly out of balance, the first move is straight up.  Vertical acceleration.  Downward pressure on the H stab will begin to pitch the nose up, increasing AOA. Any momentary DECREASE in AOA is trivial, the tail downforce will be pushing the nose up. The increase in AoA will move center of lift FORWARD in the wing, further pitching nose up.  With no pilot control inputs, the airspeed will decrease, the plane reaches the top if its ascent, and will then pitch down and recover as it "becomes one" with the air mass and is no longer being accelerated by it.
Trim condition is also very important here. One of the holy grails in free flight is to have the plane rise and fall with a gust without excessive pitching.  This is accomplished by avoiding CG placement too far forward, which requires much more elevator down trim.
This is one case where having CG further back and less elevator trim is beneficial.  As long as the aircraft is directionally stable, there is no need for having CG forward by rote!
So, suddenly reducing weight will lead to a rise/ pitch up/ airspeed reduction/ pitch down/ recovery cycle in a properly designed and trimmed aircraft, along with considerable relief for those carrying out the task.

After pitch stabilizes, aircraft will climb unless throttle is reduced.

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Why do the Cirrus SR-20 and SR-22 have the CAPS (parachute) system?
Cirrus touts the parachute as an added layer of safety. The parachute has obvious marketing value, but is the parachute's added safety a spin on the airframe's shortcomings? CAPS is not optional when buying a Cirrus, so what's the real reason for it?
Why do the Cirrus SR-20 and SR-22 have the CAPS parachute? Can a Cirrus SR-20 or SR-22 be certified to fly without it?

"Why?" is a pretty vague question.   The only appropriate question you asked is, "Can the airplane even get certified to fly without it?".

I agree, so I clarified it. I'm feeling good about the question. Even gave myself a pat on the back earlier.

I hate to break the news to you after the pat on the back and everything, but this is actually a dupe.

Mark it. I just reversed the pat. Tough day after all.

@reirab I have reinstated my back pat this morning. That other question doesn't answer the part about whether or not the Cirrus can be certified without CAPS.

Yeah, after reading both questions more closely, I decided I liked yours better and, thus, didn't vote to close. :) It might be good to link the other one as related to this question, since it does have a couple of good, related answers, then I can clean up my comments.

The FAA did not require the chute but Cirrus made it an airworthiness requirement , hence a legal operating chute is required  to maintain an airworthiness certificate.

@JohnHoland read the accepted answer, it wouldn't meet the spin requirements of 23.221

Briefly, without the CAPS the aircraft could not be certified because it wouldn't meet the spin recovery requirements of 14 CFR 23.221.
You can read a lot of detail in Cirrus's own CAPS Guide but their design premise is that pilots are bad at recovering from spins, especially close to ground. Cirrus wanted to make a safer aircraft that could recover even if the pilot has no spin training at all, so they implemented two safety features: a "cuffed wing design" from NASA and the CAPS.
Normally, the Cirrus would have to comply with 14 CFR 23.221(a):

Normal category airplanes. A single-engine, normal category airplane must be able to recover from a one-turn spin or a three-second spin,
  whichever takes longer, in not more than one additional turn after
  initiation of the first control action for recovery, or demonstrate
  compliance with the optional spin resistant requirements of this
  section.

But, rather than comply directly by demonstrating the spin recovery, Cirrus asked the FAA to certify them based on the wing design and CAPS instead, as the CAPS Guide explains:

Given that Cirrus had demonstrated enhanced low speed handling
  characteristics that will help pilots to  avoid inadvertent spin entry
  and the presence of CAPS, the FAA granted Cirrus an Equivalent Level
  of  Safety (ELOS) for the spin recovery requirement of the
  certification regulations. This ELOS is accepted  by all civil
  aviation authorities that have certified the Cirrus SR20 and SR22

The CAPS also has some other benefits, such as being easy to use for an untrained passenger (think pilot incapacitation). That might be an important "why" reason and selling point from a Marketing perspective but it isn't directly relevant for certification.

because it wouldn't meet the spin recovery requirements - do we know that it would not meet the requirements if tested, or did Cirrus simply ask for the parachute in lieu of spin certification?

@SteveV. According to a Cirrus test engineer: "The upside of the cuffs – that they give greater control in the slow flight regime – also have, like every aircraft feature, a compromise, and that is that they can cause an aircraft to take more than the required one-turn recovery rotation to be eligible to be spin-certified". You can read the full thing but basically CAPS is the only reliable spin recovery method in a Cirrus.

"A single-engine, normal category airplane must be able to recover from a one-turn spin or a three-second spin, whichever takes longer" Holy cow. Can you imagine a one-turn spin taking less than three seconds? I do not want to be in an airplane that is spinning more than 120 degrees/second, even if it does have an airframe parachute.

@Pondlife - Terrific link, thanks very much for that!

There is as always in the aviation industry a more original reason for CAPS in a Cirrus. Alan Klapmeier suffered a mid-air collision in 1985 which he survived with part of his wing gone, while the other participant in the collision died. That incident motivated him to make aviation as safe as possible, hence the chute.

That’s incorrect.  The Cirrus WILL recover from a spin and can be certified  per 23.221.  I know this from a Cirrus Test pilot in Duluth who has spun these airplanes and they will recover.  The CIRRUS SR POHs indicate the use of the CAPS system for spins both for ease of recovery and legal CYA.

@CarloFelicione I didn't say the Cirrus is completely unrecoverable from a spin, I said it's only reliably recoverable with the chute. That's the official Cirrus position from their own CAPS Guide: "Any time a Cirrus pilot experiences a
loss of control or spin, the use of CAPS is required." It's also what COPA teaches in their type training, and I've read a lot of articles saying that encouraging pilots to pull the chute has led to a decrease in fatal accidents. There's a huge difference between what a test pilot can recover from, and what an average pilot can.

A Cirrus is not spin.Certified, just like a 172/182 etc.  the chute was part of the original design, not added to get certified.  The majority of GA planes are also NOT spin certified.

There is no shortcoming to the Cirrus SR airframes requiring the installation of a ballistic parachute to make them safe for spins.  As I commented, the Cirrus SR aircraft have been spun and recovered successfully during flight testing, though the spin certification process was bypassed with the wing cuff technology to provide more roll authority and stall prevention.
Cirrus originally added the parachutes an an integral part of every new airplane which they designed as a result of a mid air collision Cirrus founder Alan Klapmeier was involved in.  He resolved to find a means to make light aircraft safer for their uses and found the best solution lay in the installation of ballistic parachutes.
The POH and CAPS training documents list activiation of the CAPS in the event of a spin both for ease of recovery and to make the OEM more immune to litigation.
The rumors that Cirruses (Cirrii?) cannot recover from spins fester on GA forums by ignorant pilots trash talking each other about their aircraft.  Other aircraft such as the Lancair/Columbia LC41-550FG series did not undergo spin certification as part of the certification process and were not equipped with ballistic parachutes (they also can successfully recover from spins as well).

"There is no shortcoming of the Cirrus SR airframes..." Are you asserting an opinion that the inability of the airframe to meet the requirements of 14 CFR 23.221 does not constitute a "shortcoming?" Or are you stating that they actually can meet the requirements of 14 CFR 23.221?

They can but circumvent those requirements, and it has nothing to do with the CAPS system.

This Cirrus engineer says the airframe can't meet the spin requirements. http://www.kineticlearning.com/pilots_world/safety/06_05/article_06_03.html

FALSE:  it just states that the entire spin certification process was not completed.  This is not proof that the airplane cannot successful do so or recover from a spin.   Rather it affirms my argument that Cirrus simply decided to circumvent the spin test requirements using the wing cuff design and a spin awareness and prevention philosophy.

"they can cause an aircraft to take more than the required one-turn recovery rotation to be eligible to be spin-certified"

Stress “can”, not “does”.  And the engineer adds the caveat “But people should understand that even though an airplane may be spin certified, certain ‘cliffs,’ as we call them, can still exist in spin aerodynamics making recovery uncertain.“. So certification does not necessarily mean certainty of recovery in all situations as well.

My father has a fairly detailed answer to this question, based on experience.  His particular evaluation of the parachute:

Don’t fly a single engine plane that isn’t equipped with a parachute.
  Although the chances of actually encountering an emergency situation
  that is worthy of “pulling the chute” are probably small to
  infinitesimal over the course of any given pilot’s career, the penalty
  for not having a parachute is almost certain death. Each pilot has to
  establish and evaluate their own risk assessment criteria, but for me
  something that has a greater than 50% risk of death, even if only 1%
  of the time, is an unacceptable risk. That’s why I bought a Cirrus in
  the first place.

Link only answers are discouraged because the link may go dead at a later date.  Please include the relevant parts of the link in your answer, and I'll be happy to switch my down to an up.

@SteveV. Fixed, my apologies.

I don't want to criticize your father, but his comment that making an off-airport landing is "almost certain death" is certainly not the case. AOPA says the fatality rate is 10%, or 20% for ditching. That's a lot, and definitely not something to take lightly, but it's far from "almost certain death". On the other hand, your father experienced a seizure and in that specific scenario a chute might be much more useful than in a 'typical' engine failure case where the pilot is still fully alert and in control.

@Pondlife No offense taken.  It's possible that his assessment of "an emergency situation that is worth of 'pulling the chute'" is more severe than simply an off-airport landing, maybe more similar to his own situation, where recovery is impossible or nearly so.  I can't speak for him, though.  It does sound like he's saying that the chute provides absolute safety in circumstances where previously there was absolute risk -- a valid point, not unlike airbags in cars.

But if he had a seizure, a second engine wouldn't save him, so why does he mention 'single engine'? Parachute on single engine airplanes, as opposed to all others, makes a difference only in case of an engine failure.

@Pondlife a chute would only be useful in that case if the pilot was aware that he was going to have a seizure far enough in advance to activate the chute, right?

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text:
Do any airplane designs exist that don't involve a flight surface that provides downforce?
Most aircraft maintain longitudinal stability by balancing three forces:

The down force acting through the center of gravity (CG)
The lifting force acting through the center of lift
The down force acting through the center of pressure on the horizontal stabilizer.

My question is, what are some aircraft designs that deviate from this basic principle? Are there aircraft we see all the time that don't work this way? As a side note, how would weight and balance be affected if this principle isn't used?

What makes you think that most aircraft are trimmed for downforce on the tail in level flight?

It is valid to choose either the CG or the Center of Lift or any other arbitrary point as the pivot point for a torque balance calculation. This question chooses the Center of Lift as the pivot point. It needs to be noted that the Center of Lift is not fixed-- it moves fore and aft-- so while the force applied at the CG is constant (at least in 1-G flight), the resulting nose-down torque about the Center of Lift is not constant. So the situation is much more complex than the diagram implies. The question could be improved but care would be need to be taken not to invalidate existing answers.

There are multiple configurations which are possible with tail or canard, which based on their locations and whether they produce lift or down-force, results in a stable or unstable aircraft (taking the aircraft center of gravity into account). The figures below show some of the possible configurations.

Source: f-16.net
In the most general case, there is no need for even a canard or a horizontal tail as long as a (down or up) force is produced in some way. A good example is the Boeing Bird of Prey, where the chines produced (vortex) lift and thus a stabilizing force. The forces acting on the longitudinal axis of the aircraft look something like this, leading to a statically stable aircraft without any lifting surface except the wing.

Profile from bj-o23.deviantart.com, others own work.

Where is the case of stable wing-tail with lifting tail?  See https://aviation.stackexchange.com/questions/22087/do-any-airplane-designs-exist-that-dont-involve-a-flight-surface-that-provides/76837#76837

Also I believe there could be a case of a stable canard where the canard is lifting or neutral, not creating a downforce, as long as the canard is meeting the air at less positive angle-of-attack than the main wing. –

Of course, just put the center of gravity back to its rear limit and fly slowly. Then all of them will produce positive lift on their tails.
Stability is not produced by a downforce at the tail. The newest book I read which claimed this was from 1911 (I happened to read the 1913 edition). Stability is produced by making the lift per area of the forward parts higher than that of the rear parts. If the lift on the forward wing is high enough, the rear wing can provide stability with a positive lift.
Canards should be the easiest to prove this: Their main wing produces positive lift, and still they can be made to fly stable.
If you are looking for a design which seems to deviate from this basic principle, maybe the Fauvel AV.36 will do:

Fauvel AV.36 in flight (picture by Daniel-Wales-Images)
Charles Fauvel designed several flying wing gliders, of which the AV.36 is the most popular. All of them had positive, natural pitch stability. There is only one wing, and it must produce lift, or the plane couldn't fly, right?
The only way to really do away with this principle is to give up natural longitudinal stability. Weight and balance would be affected such that the center of gravity is behind the neutral point, where the angle-of-attack-dependent lift force acts. This kind of airplane, however, would be very hard to fly manually.

Is this practical and has it been done either experimentally or in production? Though I guess the question is "does a design exist"...

I'm going to do some research on the airplanes I fly to see if the most rearward allowable CG position gets far enough back for the tail to be required to generate upward lift. It seems doubtful that it's allowed to get that far back for normal category flight.

@RyanBurnette I'm fairly sure the center of lift in the PA28 I fly is close to station 95.5 (the rearward CG limit, give or take somewhere around the fuel filler cap) in slow flight -- you could probably fly with zero tail downforce, and maybe even need some upward force if you really hang it on the prop. (Getting your CG that far aft however is left as an exercise for the reader: I expect it would involve adding a good quantity of ballast into the aftmost part of the tail)

@voretaq7 I once hung a lead cannonball-shaped fishing weight (50 pounds as best as I recall) off the tail tie-down eyelet of a Cessna 152 for some CG experiments.  Had additional 20-pound bags of lead that I could re-position within cabin by pulling on strings.  Verified that takeoff CG was within limits of course but in-flight re-positioning did achieve completely neutral pitch stability (untrimmable).

A tandem wing airplane has two sets of wings, each providing upward lift. One is near the front of the plane, one is near the back, and the center of gravity is between them.

(source: nurflugel.com)
The Rutan Quickie is one such plane:

(souce: wikimedia)
This design is actually fairly old; it even predates (successful) heavier-than-air flight, as it was used on the Langley Aerodrome, an experimental attempt at manned flight in the 1890s.

This would be great answer if you expanded it to include some actual examples (Rutan Quickie comes to mind).

@JanHudec Confession: I only really know the tandem wing is a thing, I don't actually know much about it. :) I'd be happy to mention the Rutan Quickie, though other than that I can only really crib examples from the wiki page. Is that acceptable behavior?

. . . First thoughts seeing that thing: "Yup, that's a Burt Rutan design alright!"

This answer will approach the question by focusing on a more narrow case:
"In a conventional wing + tail configuration, is it possible for an aircraft to be stable even if the tail is creating an upforce rather than a downforce?"
The Center of Lift (or Center of Pressure) is the point where we can treat the wing's lift vector as acting, without having to also apply an additional pitching moment to account for a cambered airfoil's inherent nose-down pitching tendency.
The Center of Lift is not fixed-- a point that is not obvious from the diagram in the original question.
If we choose to use the Center of Lift as the pivot point in a pitch torque calculation (which is a valid choice), we are relieved of having to directly account for the wing's lift force in our calculation, but we do have to recognize that the Weight vector, while constant in magnitude (at least if we confine our calculations to the 1-G case), does not act at a fixed distance from the Center of Lift, and thus the nose-down pitching moment due to Weight is not constant.  Therefore the situation is much more complex than the diagram in the original question implies.
The Center of Lift moves forward as we increase angle-of-attack and moves aft as we decrease the angle-of-attack.  The Center of Lift can even be located beyond the physical edges of the wing itself.
This makes it more intuitive as well as more convenient to treat the wing's lift vector as acting at a fixed point called the Aerodynamic Center of the wing, while also applying an additional nose-down pitching moment which is airspeed-dependent (i.e. the pitching moment coefficient is fixed but the resulting nose-down pitching moment increases as airspeed increases.)
But since the original question takes the Center of Lift approach, we'll do the same in this answer.
Since the Center of Lift migrates forward as we increase angle-of-attack and aft as we decrease angle-of-attack, the aircraft will tend to pitch up as we slow down and will tend to pitch down as we speed up.  This is destabilizing-- it is the reason that we need a horizontal tail.  But must the tail always create a downforce in order to allow stable flight?
At any given airspeed, if the CG is behind Center of Lift of the wing, the tail must create an upforce rather than a downforce.  Since the Center of Lift moves forward as we increase angle-of-attack, it follows that as we experiment with increasingly aft CG locations, we'll see the tail create an upforce at low airspeed before we see the tail create an upforce at high airspeed.
The Center of Lift of the wing should not be confused with the aircraft's Neutral Point.  The Neutral Point is the point where the aircraft has no static pitch stability.  If the CG is ahead of the Neutral Point, the aircraft is statically stable in pitch, and if the CG is aft of the Neutral Point, the aircraft has statically unstable in pitch.  The latter case is basically only the realm of computer-stabilized fly-by-wire aircraft that sacrifice stability for enhance maneuverability or enhanced cruise efficiency.
In contrast, many aircraft have lifting tails.  For example, many vintage "free flight" model airplanes have large, positively cambered tails.  (See for example this image.)  Obviously these aircraft cannot be statically unstable in pitch.  The CG must still be ahead of the Neutral Point, even though it is behind the wing's Center of Lift.
An argument can be made that a large lifting tail is generally not as efficient as a small neutrally-lifting or slightly down-lifting tail.  That argument is beyond the scope of this answer, so we should save further exploration of that for another ASE question.
Also, plenty of aircraft have a tandem wing configuration, in which the tail is actually a large lifting wing.
Here is a good description of an experiment that verified that the horizontal tail was creating an upforce in flight with a Cessna 172 with a CG that was within, but near the aft edge, of the prescribed envelope -- see last two paragraphs of section-- https://www.av8n.com/how/htm/aoastab.html#sec-pitch-equilibrium
Here is another ASE answer that gives a good treatment of the difference between Center of Pressure (or Center of Lift), Aerodynamic Center, and Neutral Point-- What is the difference between centre of pressure, aerodynamic centre and neutral point?
In summation, the Neutral Point is not the point where the aircraft will balance with the tail creating no upforce or downforce.  Rather, the Center of Lift or Center of Pressure of the wing (or to be as precise as possible, the Center of Pressure of the entire aircraft other than the tail, including effects from offset thrustlines, drag from floats, etc) is the point where the aircraft will balance with the tail creating no upforce or downforce.  It is very possible for the CG to be behind the wing's Center of Lift (or the entire aircraft's Center of Pressure) but ahead of the aircraft's Neutral Point, thus causing the tail to generate an upforce, yet still allowing positive static pitch stability.
The basic requirement for static pitch stability is simply that the tail is flying at a less positive angle-of-attack, or at least a less positive lift coefficient, than the wing.  This includes, but is not confined to, cases where the tail is flying at a negative angle-of-attack and creating a downforce.  You can read more about it in this section of an outside website-- https://www.av8n.com/how/htm/aoastab.html#sec-teeter .
This related ASE answer is also enlightening-- Do any airplane designs exist that don't involve a flight surface that provides downforce?

Obviously adding floats affects the location of the neutral point (and changes the stabilizer trim required to hold a given angle-of-attack of the wing)--

(future edit-- could note that first linked ASE answer has a diagram that assumes tail is uplifting not downlifting)

possible future edit-- remove quotes from second sentence

possible future edit-- introduce the idea of the center of pressure of the entire aircraft (excluding the tail) earlier in the answer.